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I'm actually an engineering student so I'm not too good with probability and was hoping someone may be able to help with the following:

So I have a ratio of discrete random variables. I want to be able to know when I can get away with approximating its expected value as a ratio of expected values. For example, for the continuous case I came across the following formulation from a Taylor series expansion

$\text{E}[x/y]=\text{E}[x]/ \text{E}[y]- \text{Cov}[x ,y]/ \text{E}[y]^2 + \text{E}[x] \text{Var}[y]/ \text{E}[y]^3$

Does this apply for discrete random variables? if not, what is its analogous expression?

thank you! hadi

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How did you derive the formula? Thanks! –  qed Sep 5 '13 at 15:38
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3 Answers

A simple, general, good approximation purely in terms of expectations does not seem achievable. For example, let $\epsilon$ be a very small positive number, and let $X$ and $Y$ be independent, each taking on the value $\epsilon$ with probability $1/2$, and $1$ with probability $1/2$. Then $E(X/Y)$ is very large, but the various moments are all quite reasonable.

Possibly some reasonable approximations are possible if all values of $Y$ are well away from $0$.

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That last term doesn't make sense to me. Let $E[x] = \mu_x$ and $E[y] = \mu_y$. From the Taylor series of $x/t$ around $t=\mu_y$, $x/y = x/\mu_y - x(y -\mu_y)/\mu_y^2 + x (y - \mu_y)^2/\mu_y^3 + \ldots $ resulting in $E[x/y] = \mu_x/\mu_y - {\rm Cov}[x,y]/\mu_y^2 + E[x (y - \mu_y)^2]/\mu_y^3 + \ldots$, but $E[x (y - \mu_y)^2] \ne \mu_x {\rm Var}[y]$ unless $x$ and $(y - \mu_y)^2$ are uncorrelated.

In any case, it has nothing to do with discrete vs. continuous: it is just as valid for discrete random variables as it is for continuous ones. The main point is that to have a good approximation (apart from that last problematic term) you'll want $y - E[y]$ to be (with high probability) small compared to $|E[y]|$. You especially need to avoid $y$ near 0, which could make $E[x/y]$ be undefined.

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First of all, $X$ and $Y$ should be independent. Now the problem is that in general $E[1/y] \neq 1/E[y]$. However, this is a reasonable approximation when $y$ is concentrated around some "large" value.

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